Underdamped Oscillator Solution: Deriving x(0) and v(0)

AI Thread Summary
The discussion focuses on deriving the underdamped oscillator solution, specifically showing that x(0) equals x₀ and v(0) equals v₀. The proposed solution is x(t)=x₀e^{-γt}[cos(Ω't+((v₀+γx₀)/(x₀Ω')sinΩ't]. Participants express confusion about the transformation of the general form of the oscillator equation and seek clarification on the correct approach. One contributor suggests that the equation could be simplified further but acknowledges their understanding is limited. The conversation highlights the need for precise mathematical manipulation to validate the initial conditions of the oscillator.
yakkayakka
Messages
1
Reaction score
0

Homework Statement




Show that the underdamped oscillator solution can be expressed as x(t)=x_{0}e^{-γt}[cos(Ω't+((v_{o}+γx_{o})/(x_{o}Ω')sinΩ't] and demonstrate by direct calculation that x(0)=x_{o} and \dot{x}(0)=v_{o}

Homework Equations



The underdamped oscillator solution is
x(t)=ae^{-γt}cos(Ω't+\alpha)

The Attempt at a Solution


This problem completely overwhelms me so my solution may be a little lacking...
I took the general form
Acos(ω_{o}t)+Bsin(ω_{o}t)
Where
A=acos(\alpha) and B=-asin(\alpha)
Which according to what I read in the book should yield
x(t)=a[cos(ω_{o}t+\alpha)]
So I am thinking that the equation ae^{-γt}cos(Ω't+\alpha) can be transformed into a more useful form using the same method

and that is sadly as close as I could get

Any input would be appreciated. Thanks.
 
Physics news on Phys.org
You wrote,

x(t)=x0e−γt[cos(Ω't+((vo+γxo)/(xoΩ')sinΩ't]

I think you are missing some ")" somewhere?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top